Playing with idempotents in the ring of Witt vectors

DEFINITION 9.3   Let $A$ be a commutative ring. For any $a\in A$ , we denote by $[a]$ the element of $\mathcal W_1(A)$ defined as follows:

$$[a]=(1-a T)_W$$

We call $[a]$ the Teichmüller lift'' of $a$

LEMMA 9.4   Let $A$ be a commutative ring. Then:

1. $\mathcal W_1(A)$ is a commutative ring with the zero element $[0]$ and the unity $[1]$ .

2. For any $a,b \in A$ , we have
   $$[a]\cdot [b]=[a b]$$

$\qedsymbol$

PROPOSITION 9.5   Let $p$ be a prime number. Let $A$ be a ring of characteristic $p$ . Then:

1. If $n$ is a positive integer which is not divisible by $p$ , then $n$ is invertible in $\mathcal W_1(A)$ . To be more precise, we have
   $$\frac{1}{n}\cdot [1]= \left( (1-T)^{\frac{1}{n}}\right)_W = \left((1+\sum_{j=1}^\infty \binom{\frac{1}{n}}{j} (-T)^j \right)_W.$$

2. $p \cdot : \mathcal W_1(A)\to \mathcal W_1(A)$ is an injection.
3. For any positive integer $n$ which is not divisible by $p$ , we define an element $e_n$ as follows:
   $$e_n=\frac{1}{n}\cdot (1-T^n)_W.$$
   Then:
   1. For any positive integer $n$ , $e_n$ is an idempotent.
   2. If $n\vert m$ , then $e_n \succeq e_m$ in the order of idempotents.

PROOF.. (1) follows from the next lemma. The rest is easy. $\qedsymbol$

LEMMA 9.6   Let $n$ be a positive integer. Let $k$ be a non negative integer. Then we have always

$$\binom{\frac{1}{n}}{k}\in \mathbb{Z}\left[\frac{1}{n}\right].$$

PROOF..

$$\binom{\frac{1}{n}}{k}\in \mathbb{Z}\left[\frac{1}{n}\right]$$

$$= \frac{\frac{1}{n}(\frac{1}{n}-1)\cdots (\frac{1}{n}-(k-1))}{k!}$$

$$= \frac{1}{n^k} \frac{(1(1-n)(1-2n)\dots (1-(k-1)n)}{k!}$$

So the result follows from the next sublemma. $\qedsymbol$

SUBLEMMA 9.7   Let $n$ be a positive integer. Let $k$ be a non negative integer. Let $\{a_j\}_{j=1}^k\subset \mathbb{Z}$ be an arithmetic progression of common difference $n$ . Then:

1. For any positive integer $m$ which is relatively prime to $n$ , we have
   $$\char93 \{j;\ m \vert a_j\ \} \geq \left\lfloor \frac{k}{m} \right\rfloor$$

2. For any prime $p$ which does not divide $n$ , let us define
   $$c_{k,p}= \sum_{i=1}^\infty \lfloor \frac{k}{p^i}\rfloor$$
   (which is evidently a finite sum in practice.) Then
   $$p^{c_{k,p}} \vert \prod_{j=1}^k a_j$$

3. $$p^{c_{k,p}}\vert k!,\qquad p^{c_{k,p}+1} \nmid k!$$

4. $$\frac{\prod_{j=1}^k a_j }{k!} \in \mathbb{Z}_{(p)}$$

PROOF.. (1) Let us put $t=\lfloor\frac{k}{m}\rfloor$ . Then we divide the set of first $kt$ -terms of the sequence $\{a_j\}$ into disjoint sets in the following way.

$$S_0=\{a_1,a_2,\dots, a_m\},$$

$$S_1=\{a_{m+1},a_{m+2},a_{m+m}\},$$

$$S_2=\{a_{2m+1},a_{2m+2},a_{2m+m}\},$$

$$\dots$$

$$S_{t-1}=\{a_{(t-1)m+1},a_{(t-1)m+2},\dots, a_{(t-1)m+m}\}$$

Since $m$ is coprime to $n$ , we see that each of the $S_u$ gives a complete representative of $\mathbb{Z}/n\mathbb{Z}$ .

(2): Apply (1) to the cases where $m=p,p^2,p^3,\dots$ and count the powers of $p$ which appear in $\prod a_j$ .

(3): Easy. (4) is a direct consequence of (2),(3). $\qedsymbol$

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